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Old Jul 14th 2011, 04:37 PM   #1
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Quick Circuit Question, Direction of Flow

Hi all, I've got a quick question about current flow in a circuit. Our prof told us that current flows out of the positive end of the battery, but he didn't tell us how it works in a circuit with 2 batteries. I couldn't figure it out from the text, either, and I haven't found the answer online.

Here's the problem in question: http://i.imgur.com/EbU3I.png. You can see pencil-smears from where I was trying to determine the flow. I know that current is flowing from the positive end of the 12-Volt, because I solved for Amps through the 3.9-Ohm resistor and the back of the book said my answer was right (the problem does not mention direction, and was asking for the Amps at each resistor). Using the same logic for the 9-Volt, however, got me the wrong Amperage for the 6.7-Ohm resistor (book says it should be 1.0A). How can I figure out current flow without looking in the back of the book?
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Old Jul 18th 2011, 01:15 AM   #2
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If you want to understand it better, you might want to learn about Kirchhoff's laws.
Kirchhoff's circuit laws - Wikipedia, the free encyclopedia

Otherwise, your arrows seem fine for me. Current goes from the two positive teminals, then add (merge) at the first junction, and separate again at the second junction to go back to the negative terminals.

Let's name the current from the 12V cell as I1 (red) and the current from the 9V cell I2 (blue).



Now, taking the first loop, (where I1 goes), we get (we are finding the potential differences across each component, and that's Kirchhoff's Voltage Law):
12 = 3.9I1 + 1.2(I1+I2) + 9.8I1

I take all the power supplies on the left, the P.d is 12, then from V = IR, I get the P.d across each resistor. The I1+I2 is from Kirchhoff's Current law. The two currents add up.

Taking the second loop now, that is where the blue current flows;
9 = 6.7I2 + 1.2(I1+I2)

And taking a third loop, as the one without the central wire,
12 - 9 = 3.9I1 - 6.7I2 + 9.8I1

Okay, couple of comments here. I took the direction as following the red current ONLY. That's because the 12V is larger than 9V. That way, the 9V cell is going against the current I imposed. If ever you are not sure about which potential is higher, just assume one direction, but you must stick to it through the whole loop. Your answer will not be affected.

Next, notice I used -6.7I2, this is because the current I drew is going in the opposing direction! With those three equations now, you should be able to get the answers
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